Catching up on e-mails ...as usual :-(

and I noticed no one had answered this question.
The ORA-3232 is usually not related to tablespaces
but to the relationship between the size of your
temp tablespace INITIAL and NEXT extent in relation
to sort_area_size (and hash joins).

This hit me not too long ago when I had a really
small temp extent size for a sandbox I was planning with.
For me, I just created my sandbox temp space with larger
initial extent. There are also Metalink Notes about
changing the HASH_MULTIBLOCK_IO_COUNT to resolve this.

When you "sort_area_size=20M", all sorts were done in
memory so you didn't trigger this condition.

- Better late than never
Babette

-----Original Message-----
George
Sent: Thursday, November 07, 2002 10:59 AM
To: Multiple recipients of list ORACLE-L



The following query is causing the following error

ERROR at line 1:
ORA-03232: unable to allocate an extent of 22 blocks from tablespace 3

select count(l.processid) from tmslog l, tmslogtimeout t where l.processid =
t.processid and l.statifiedflag='Y' and t.processcompleteflag='Y'

Tablespace #3 is temp, 800 MB, 128K extent size locally managed. The user is
also set to use temp.

If I do a alter session set sort_area_size=20M then it completes. Currently
the sort_area_size is set via the init file as 5 mb.

Ideas ?

George
________________________________________________
George Leonard
Oracle Database Administrator
Dimension Data (Pty) Ltd
(Reg. No. 1987/006597/07)
Tel: (+27 11) 575 0573
Fax: (+27 11) 576 0573
E-mail:[EMAIL PROTECTED]
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You Have The Obligation to Inform One Honestly of the risk, And As a Person
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Once Informed & Totally Aware of the Risk, Every Fool Has the Right to Kill
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-----Original Message-----
Sent: 07 November 2002 15:24 PM
To: Multiple recipients of list ORACLE-L

Thanks Kevin, good to hear from you. As usual you're Johnie on spot with
TFM. It's interesting that this can be overcome with the inline view
technique posted earlier by Raj.


Steve


-----Original Message-----
Sent: Wednesday, November 06, 2002 5:23 AM
To: Multiple recipients of list ORACLE-L


Directly from TFM....

Notes on Hierarchical Queries:

If you specify a hierarchical query and also specify the ORDER BY clause,
the ORDER BY clause takes precedence over any ordering specified by the
hierarchical query, unless you specify the SIBLINGS keyword in the ORDER BY
clause.

The manner in which Oracle processes a WHERE clause (if any) in a
hierarchical query depends on whether the WHERE clause contains a join:

    * If the WHERE predicate contains a join, Oracle applies the join
predicates before doing the CONNECT BY processing.

    * Oracle applies any non-join predicates (that is, all predicates if the
WHERE clause does not contain a join) after doing the CONNECT BY processing
without affecting the other rows of the hierarchy.

-----Original Message-----
Sent: Tuesday, November 05, 2002 3:29 PM
To: Multiple recipients of list ORACLE-L


I was also able to confirm this works on O9i.

-----Original Message-----
Sent: Tuesday, November 05, 2002 11:14 AM
To: Multiple recipients of list ORACLE-L


I get an error on 8.1.7.2. Is "siblings" new?

SQL> l
  1  SELECT LEVEL, treenode.*
  2    FROM treenode
  3   START WITH parentid=0
  4  CONNECT BY PRIOR ID = parentid
  5* ORDER SIBLINGS BY PARENTid , nodeorder
SQL> /
ORDER SIBLINGS BY PARENTid , nodeorder
      *
ERROR at line 5:
ORA-00924: missing BY keyword


-----Original Message-----
Sent: Tuesday, November 05, 2002 11:02 AM
To: '[EMAIL PROTECTED]'; Orr, Steve


SELECT LEVEL, treenode.*
  FROM treenode
 START WITH parentid=0
CONNECT BY PRIOR ID = parentid
ORDER SIBLINGS BY PARENTid , nodeorder
Raj
______________________________________________________
Rajendra Jamadagni              MIS, ESPN Inc.
Rajendra dot Jamadagni at ESPN dot com
Any opinion expressed here is personal and doesn't reflect that of ESPN Inc.

QOTD: Any clod can have facts, but having an opinion is an art!


-----Original Message-----
Sent: Tuesday, November 05, 2002 12:24 PM
To: Multiple recipients of list ORACLE-L


Challenge: present SQL results hierarchically and sort the nodes. Use sort
column without changing data. Here's the DDL/DML to start:
create table treenode (
        id              number          not null
                        constraint pk_treenode primary key,
        parentid        number          not null,
        nodeorder       number          not null,
        description     varchar2(20)    null);
insert into treenode values(1,0,0,'top folder');
insert into treenode values(9,1,0,'1st subfolder');
insert into treenode values(7,1,2,'3rd subfolder');
insert into treenode values(2,1,1,'2nd subfolder');
insert into treenode values(8,7,1,'folder 3 item 2');
insert into treenode values(6,2,3,'folder 2 item 3');
insert into treenode values(5,7,0,'folder 3 item 1');
insert into treenode values(3,2,2,'folder 2 item 2');
insert into treenode values(4,2,1,'folder 2 item 1');
-----------------------------------------------------
Here's the data presented hierachically without the desired sort:
select * from treenode
start with parentid=0 connect by prior id = parentid;
        ID   PARENTID  NODEORDER DESCRIPTION
---------- ---------- ---------- --------------------
         1          0          0 top folder
         9          1          0 1st subfolder
         7          1          2 3rd subfolder
         8          7          1 folder 3 item 2
         5          7          0 folder 3 item 1
         2          1          1 2nd subfolder
         6          2          3 folder 2 item 3
         3          2          2 folder 2 item 2
         4          2          1 folder 2 item 1
-----------------------------------------------------
Desired SQL statement results:
        ID   PARENTID  NODEORDER DESCRIPTION
---------- ---------- ---------- --------------------
         1          0          0 top folder
         9          1          0 1st subfolder
         2          1          1 2nd subfolder
         4          2          1 folder 2 item 1
         3          2          2 folder 2 item 2
         6          2          3 folder 2 item 3
         7          1          2 3rd subfolder
         5          7          0 folder 3 item 1
         8          7          1 folder 3 item 2
-----------------------------------------------------
Kudos to anyone who can figure out how to do this via SQL.


Steve Orr
Bozeman, Montana
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